GaI3 Lewis structure

GaI₃ [gallium(Ⅲ) iodide] has one gallium atom and three iodine atoms.

In GaI₃ Lewis structure, there are three single bonds around the gallium atom, with three iodine atoms attached to it, and on each iodine atom, there are three lone pairs.

Alternative method: Lewis structure of GaI₃

Rough sketch

• First, determine the total number of valence electrons

In the periodic table, gallium lies in group 13, and iodine lies in group 17.

Hence, gallium has three valence electrons and iodine has seven valence electrons.

Since GaI₃ has one gallium atom and three iodine atoms, so…

Valence electrons of one gallium atom = 3 × 1 = 3
Valence electrons of three iodine atoms = 7 × 3 = 21

And the total valence electrons = 3 + 21 = 24

Learn how to find: Gallium valence electrons

• Second, find the total electron pairs

We have a total of 24 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 24 ÷ 2 = 12

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since gallium is less electronegative than iodine, assume that the central atom is gallium.

Therefore, place gallium in the center and iodines on either side.

• And finally, draw the rough sketch

Lone pair

Here, we have a total of 12 electron pairs. And three Ga — I bonds are already marked. So we have to only mark the remaining nine electron pairs as lone pairs on the sketch.

Also remember that gallium is a period 4 element, so it can keep more than 8 electrons in its last shell. And iodine is a period 5 element, so it can keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are iodines.

So for each iodine, there are three lone pairs, and for gallium, there is zero lone pair because all nine electron pairs are over.

Mark the lone pairs on the sketch as follows:

Formal charge

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For gallium atom, formal charge = 3 – 0 – ½ (6) = 0

For each iodine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, both gallium and iodine atoms do not have charges, so no need to mark the charges.

Final structure

The final structure of GaI₃ contains a central gallium atom linked to three iodine atoms through single covalent bonds. In this layout, the gallium atom does not complete a full octet in its monomeric form but instead maintains six valence electrons, which comprise three bonding pairs with no remaining lone pairs. Within this arrangement, each iodine atom successfully satisfies the octet rule by retaining three lone pairs alongside its single shared bond. This configuration represents the most stable state for the neutral molecule because it results in a formal charge of zero for every atom involved. Therefore, this specific electronic distribution serves as the definitive and most accurate Lewis representation of gallium triiodide.

Next: SeO₄^2- Lewis structure

External links

• https://lambdageeks.com/gai3-lewis-structure/
• https://homework.study.com/explanation/draw-the-lewis-structure-for-gai3-and-provide-the-following-information-a-number-of-bonding-electron-pairs-b-number-of-nonbonding-electron-pairs-c-electron-geometry-d-molecular-geometry-e-approximate-bond-angle.html
• https://www.bartleby.com/questions-and-answers/for-the-molecule-gai3-give-the-lewis-structure-the-valence-shell-electron-pairs-the-bonding-electron/bf8a37ed-a3e5-4c28-9b40-910420926a2e